Question

3- x2 + x + 1then value of
P(-1)+P(1) by 2

is​

Answer 1

$Let \: p(x) = 3 - x^{2} + x + 1$

$i) p(-1) \\= 3 - (-1)^{2} + (-1) + 1 \\= 3 - 1 - 1 + 1 \\= 2$

$ii) p(1) \\= 3 - 1^{2} + 1 + 1 \\= 3 - 1 + 1 + 1 \\= 4$

$\red{ Value \: of \: \frac{p(-1) + p(1)}{2} } \\= \frac{2+4}{2} \\= \frac{6}{2} \\\green { = 3 }$

Therefore.,

$\red{ Value \: of \: \frac{p(-1) + p(1)}{2} }\green { = 3 }$

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Answer 2

$\huge\bf\underline{Solution}$

Step by step explanation:-

p(x) = 3 -x² + x + 1

Value of p(-1) =

Substuite x = -1 in given equation

p(-1) = 3 - (-1)² -1 +1

p(-1) = 3 -1 -1 +1

p(-1) = 4 -2

p(-1) = 2

p(x) = 3-x² +x + 1

p(1) = Substuite x = 1

p(1) = 3-(1)² +1 +1

p(1) = 3 -1 +1 +1

p(1) = 4

So, $\frac{p(-1)+p(1)}{2}$ = $\frac{2+4}{2}$ = 3

Required answer is 3